Well, "Ramanujan summation" usually refers to something else entirely, so let's not use that phrase here. Alternate topologies on the real numbers are another red herring; they're not relevant to this discussion. We're not evaluating the limit of a sequence of partial sums in any topology. With that stuff out of the way...
There are multiple ways to compute $\zeta(-1)$. You've outlined using the functional equation to relate it to $\zeta(2)$. But the more relevant method is in the Wikipedia article:
$$-3\zeta(-1)=\eta(-1)=\lim_{x\to 1^-}\left(1-2x+3x^2-4x^3+\cdots\right)=\lim_{x\to 1^-}\frac{1}{(1+x)^2}=\frac14$$
where $\eta$ is the Dirichlet eta function, not to be confused with the unrelated and better-known Dedekind eta function.
This is a direct analogue to what Ramanujan wrote in his notebook:
$$-3c=1-2+3-4+\mathrm{\&c}=\frac{1}{(1+1)^2}=\frac14$$
Imagine, for the sake of argument, that Ramanujan's intention is in fact to compute $\zeta(-1)$. Then every single step in his computation of "$c$" corresponds to a step in the computation of $\zeta(-1)$, just with the variables $s$ and $x$ elided. And the computation of $\zeta(-1)$ is rigorous; each step is justified in the Wikipedia article, which I'll quote for completeness:
Where both Dirichlet series converge, one has the identities:
$$
\begin{alignat}{7}
\zeta(s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+\cdots& \\
2\times2^{-s}\zeta(s)&{}={}& 2\times2^{-s}&& {}+2\times4^{-s}&&{} +2\times6^{-s}+\cdots& \\
\left(1-2^{1-s}\right)\zeta(s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+\cdots&=\eta(s) \\
\end{alignat}
$$
The identity $(1-2^{1-s})\zeta(s)=\eta(s)$ continues to hold when both functions are extended by analytic continuation to include values of $s$ for which the above series diverge. Substituting $s = −1$, one gets $−3\zeta(−1) = \eta(−1)$. Now, computing $\eta(−1)$ is an easier task, as the eta function is equal to the Abel sum of its defining series, which is a one-sided limit.
And that explains why Ramanujan's manipulation yields the same value as $\zeta(-1)$.
Perhaps the most obscure step in the argument is the usage of Abel summation to evaluate a Dirichlet series. The reference to Knopp justifies that step better than I would be able to. Beyond that, I'm not sure what else you're likely to object to. Feel free to ask about any particular step!
This behavior is completely normal and is classic for Fourier/Laplace/Mellin transforms. I’ll define:
$$
f(x)=\zeta(-1/\ln x)
$$
The key is that you are not integrating over the entire domain so applying the transformation and its inverse will not give you the original function, but will set it to zero outside the domain of integration.
Take the simpler example:
$$
f(x)=1/x
$$
If you take the Mellin transform:
$$
F(s)=\int_{1/2}^1f(x)x^{s-1}dx
$$
you can compute it explicitly as:
$$
F(s)=\frac{1-2^{1-s}}{s-1}
$$
Just as in your example, $F$ is entire (removable singularity at $s=1$).
You can now consider the inverse Mellin transform:
$$
\tilde f(t)=\frac1{2\pi i}\int F(s)x^{-s}ds
$$
Indeed, since $F$ is analytic, you can take any vertical line to integrate. However, you cannot deduce that the integral is zero. Implicitly, you are assuming that you can deform the line to a circle and then contract it to get zero. This is only possible for $x\not\in(1/2,1)$ because of the essential singularities of the exponentials at infinity.
For $x<1/2$, you can deform the vertical line by closing it off at $\Re s=+\infty$. Similarly, for $x>1$, you can deform the vertical line at $\Re s=-\infty$. However, at $t\in(1/2,1)$, you are stuck as you cannot close on either side, and indeed, the value of the integral is non zero.
Furthermore, for $x\in(1/2,1)$, your contour needs to have vertical asymptotes to not “upset” the essential singularity. This means that you cannot use the integral representation to analytically continue $f$ for complex $t$ since the integral will now diverge exponentially.
This is basically what is happening with you case. For $x\not\in(1/2,1)$ you do get zero, but for $x\in(1/2,1)$ you cannot deform the vertical line contour due to the essential singularity at infinity. It’s trickier to see in your case as $K$ does not have a simple analytic expression. Also, you cannot use it to recover the meromorphic extension of $\zeta$ because the contour integral diverges as soon as $x$ is not real.
Back to my toy example, if the lower bound of the integration is taken to be $0$, the ROC is now for $\Re s>1$ and:
$$
F(s)=\frac1{s-1}
$$
You now have a pole at $s=1$. By contour integration you recover the original function $f(x)=\frac1x$ but only for $x<1$. Indeed, the same trick of closing the integral works for $x>1$.
Similarly for your case, if you chose the lower bound to be $1/e$, you’ll be able to extend the integral representation all the way to the pole at $x=1/e$.
The usual way to use Mellin type integrals for meromorphic extension is to rather reason the other way around. You need to view the function you want to continue as a Mellin transform. This is how you get the usual integral representation of the zeta function that you can then meromorphically extend.
Hope this helps.
Best Answer
There are two approaches,
one from the unique factorization in prime ideals in the PID $\Bbb{Z}[i]$ which leads to $$\sum_{(m,n)\ne (0,0)}(m^2+n^2)^{-s}=4\zeta(s)L(s,\chi_4)$$
and the other from $$\pi^{-s}\Gamma(s)\sum_{(m,n)\ne (0,0)}(m^2+n^2)^{-s}=\int_0^\infty (\theta(x)^2-1) x^{s-1}dx$$ where $\theta(x)=\sum_k e^{-\pi k^2 x}$ which is $=x^{-1/2}\theta(1/x)$ from the Poisson summation formula.
If you only care of the poles on $\Re(s)>1/2$ then the Gauss circle problem is the simplest solution.